This is a modified thought experiment about twin paradox. Twin A goes to planet X which is 10 light years away from earth at speed 0.5c, hence the journey takes 20 years. A then wait for B who goes to X afterwards at speed 0.2c, hence the journey takes 50 years.When A and B meet again on planet X, who is the older one?

If we assume that B starts after A after A reaches his destination and come to a halt, then A will age 17.32 yrs during its outbound trip and then 50 yrs waiting for B to arrive for a total of 67.32 yearsB will age 20 years during A's outbound trip and then age ~49 years during his trip for a total of ~69 years, and thus will be older than A when he arrives.

If we assume that B starts after A after A reaches his destination and come to a halt, then A will age 17.32 yrs during its outbound trip and then 50 yrs waiting for B to arrive for a total of 67.32 yearsB will age 20 years during A's outbound trip and then age ~49 years during his trip for a total of ~69 years, and thus will be older than A when he arrives.

so there is no paradox in this case. Observer staying on earth will agree with observer staying on X about their age.

If we assume that B starts after A after A reaches his destination and come to a halt, then A will age 17.32 yrs during its outbound trip and then 50 yrs waiting for B to arrive for a total of 67.32 yearsB will age 20 years during A's outbound trip and then age ~49 years during his trip for a total of ~69 years, and thus will be older than A when he arrives.

so there is no paradox in this case. Observer staying on earth will agree with observer staying on X about their age.

There is never a paradox. All observers will always agree which twin aged more if they are separated and then brought back together again, no matter what the scenario in which this occurs.

In the original paradox, the twin A reverses its velocity as soon he arrives planet X. When back to earth he aged 34,64, while the earth twin aged 40 years.

But during all his trip, earth was moving from him at 0.5c. So, from the reference frame of the traveller twin, at first glance, his brother at earth should have aged only 34,64 x 0,866 = 30 years.

But when reversing the engines at planet X, the traveller twin observed a sudden change of the clocks at earth. Just before reversing velocity, t' = 20 - 0,5 x 10 = 15 years, what should be expected, because for him, the earth was moving, and his brother at earth was aging slower (15 = 17,32 x 0,866).

But just after reversing velocity, the clock at earth shows t' = 20 + 0,5 x 10 = 25 years.

So, in the returning trip, more 17,32 passes for him, and more 15 years at earth. During all the time, the brother at earth aged slowly for the travelling brother, but the change of syncronization when reversing of the engines, made all the difference.

A consequence is that for an observer at earth, the sudden reversing of velocity of the ship would have taken 10 years. During all that time the ship (and any clock inside it) would be seen as stopped at planet X.

A consequence is that for an observer at earth, the sudden reversing of velocity of the ship would have taken 10 years. During all that time the ship (and any clock inside it) would be seen as stopped at planet X.

After I posted it, I realized it was wrong. The "news" of the arriving ship would come to earth at t = 20 + 10 = 30 years. And during the next 10 years the ship would be observed as coming back.

This is a modified thought experiment about twin paradox. Twin A goes to planet X which is 10 light years away from earth at speed 0.5c, hence the journey takes 20 years. A then wait for B who goes to X afterwards at speed 0.2c, hence the journey takes 50 years.When A and B meet again on planet X, who is the older one?

Who's still alive?

Clearly the math says a faster rate of travel incurs less age, yet who's aging more and through what time period? You'll find an answer there based on the math.

If we assume that B starts after A after A reaches his destination and come to a halt, then A will age 17.32 yrs during its outbound trip and then 50 yrs waiting for B to arrive for a total of 67.32 yearsB will age 20 years during A's outbound trip and then age ~49 years during his trip for a total of ~69 years, and thus will be older than A when he arrives.

so there is no paradox in this case. Observer staying on earth will agree with observer staying on X about their age.

There is never a paradox. All observers will always agree which twin aged more if they are separated and then brought back together again, no matter what the scenario in which this occurs.

There is no paradox from A's perspective, PROVIDED he realizes that B (she) instantaneously ages during A's instantaneous stop at his destination. If he DOESN'T realize that, then he DOES perceive a paradox when she arrives, because he would expect her to be younger at arrival. That's the SAME issue that occurs in the standard twin "paradox" scenario.

"Near" instantaneous would be better. There is nothing that prevents us with dealing with low accelerations over extended periods, it's just that for the purposes of illustration of the principles involved, it jut not worth the trouble. For example, let's consider A during that part of their outbound trip where they come to a rest with respect to B some ten light years distant from B. If he were to do this by maintaining a constant 1g acceleration( as measured by himself), then he would have to start his deceleration when he is 0.15 ly (as measured by B) from where he wants to stop. If you want to calculate how much time passes on B's clock according to A during this period, you have to factor in the fact that his relative velocity with respect to A is constantly changing, and thus the time dilation due to that. You also have to figure in the effects caused by his being in an accelerated frame. Clocks in the direction of the Acceleration (towards B) will run fast, by a factor determined by the magnitude of his acceleration and the distance to those clocks. In the case of B's clock this distance is increasing as A goes travels between the point where they first start the deceleration and where they come to a stop with respect to B. Not an impossible task, but an extra complication that doesn't really aid in the example. It is much simpler to assume that the acceleration is very high and over a very short distance. So, for example, I gave the respective ages for A and B as being 67.32 and ~69 years. But these answers are a result of rounding. As long as we assume that the acceleration phases for either A and B occurred over a very short distance any difference in the answer made by accounting for them are too small to effect the rounding.

The problem with using truly Instantaneous changes in velocity is that this equates to an infinite acceleration over zero time, and anytime you bring infinities (especially coupled with a division by zero) in, you are inviting headaches.

There is never a paradox. All observers will always agree which twin aged more if they are separated and then brought back together again, no matter what the scenario in which this occurs.

There is no paradox from A's perspective, PROVIDED he realizes that B (she) instantaneously ages during A's instantaneous stop at his destination. If he DOESN'T realize that, then he DOES perceive a paradox when she arrives, because he would expect her to be younger at arrival. That's the SAME issue that occurs in the standard twin "paradox" scenario.

I should also add that B (she) also doesn't perceive a paradox, PROVIDED she realizes that when she instantaneously changes velocity to leave home, A (he) will instantaneously age. But if she DOESN'T realize that, then she DOES perceive a paradox when she arrives at the destination, because she finds A older than she expects him to be. Again, that's the SAME issue that occurs in the standard twin "paradox" scenario.

The mathematics isn't too hard (at least for piecewise-constant accelerations), but it's not as easy as for the instantaneous velocity-change case. Instantaneous velocity changes are of course not actually possible, but it is a mistake to think that the results they provide are so unlike the realistic cases as to be of no use. There are realistic scenarios, involving acceleration segments of only 1g, that are qualitatively very similar to the instantaneous velocity-change case. On my webpage

I give an example where the traveler (he) is about 40 ly away from his home twin (according to her), traveling away from her at about .77 ly/y, and then accelerates for two years (of his life) at 1g in a direction TOWARDS her. During that acceleration, he concludes that she gets about 64 years older. So during the acceleration, she ages more than 30 times faster than he does.

If he then decides to accelerate for another two years (of his life), but this time in the direction AWAY from her, he will conclude that she gets about 59 years YOUNGER during the acceleration.

If we assume that B starts after A after A reaches his destination and come to a halt, then A will age 17.32 yrs during its outbound trip and then 50 yrs waiting for B to arrive for a total of 67.32 yearsB will age 20 years during A's outbound trip and then age ~49 years during his trip for a total of ~69 years, and thus will be older than A when he arrives.

so there is no paradox in this case. Observer staying on earth will agree with observer staying on X about their age.

There is never a paradox. All observers will always agree which twin aged more if they are separated and then brought back together again, no matter what the scenario in which this occurs.

What would happen if the calculation is done in A's reference frame?According to A, he never moves. Instead he will see B moves away at 0.5c for 20 years and then moves back at 0.2c for 50 years. Accounting for time dilation, after 70 years according to A, B would only age 17.32+48.99=66.31 years.

On the other hand, according to B, he never moves. Instead he will see A moves away at 0.5c for 20 years and then moves back at 0.2c for 50 years. Accounting for time dilation, after 70 years according to B, A would only age 17.32+48.99=66.31 years.